# Let $f(x)$ be an irreducible polynomial over $F$ of degree $n$, and let $K$ be a field extension of $F$ with $[K:F]=m$

Let $f(x)$ be an irreducible polynomial over $F$ of degree $n$, and let $K$ be a field extension of $F$ with $[K:F]=m$. If $\gcd(n,m)=1$, show that $f$ is irreducible over $K$.

I thought suppose $f$ reducible over $K$, so $$f(x) = \underbrace{p(x)}_{(\deg = r)}\underbrace{q(x)}_{(\deg = s)}$$ where $1<r,s<n$ and both divides $n$. I imagine that $r$ or $s$ divides $m$ too, but I don't know how to prove this.

Any hint? Or is there a better way?

If $f=pq$ where $p$ is irreducible in $K[x]$ with degree $r$ and $0＜r＜n$.select a root $\alpha$ of $p$ .then $[K[\alpha]:K][K:F]=[K[\alpha]:F[\alpha]][F[\alpha]:F]$,that is $rm=[K[\alpha]:F[\alpha]]n$.Hence $n\mid r$ since n and m are coprime. Contradiction
• ${}$Nice! Thanks! Commented May 4, 2018 at 6:27
• Why you assumed p is irreducible in $K[x]$? Do we need only p to be irreducible or both of them or no one of them? Commented Feb 2 at 17:53